List six positive integer solutions for each of these equations and comment on your results. Two have been done for you. x y = x + y = 4 x =, y = 0 x = 3, y = x = 4, y = x = 5, y = 3 x = 6, y = 4 x = 7, y = 5 x = 0, y = 7 x =, y = 6 x = 4, y = 5 x = 6, y = 4 x = 8, y = 3 x = 0, y = If we considered all solutions, not just positive integers, there would be an infinite number of answers. However, there is one solution which is correct for both of these equations simultaneously. teachitmaths.co.uk 03 6809
We can see this clearly on a graph: x y = x + y = 4 x =, y = 0 x = 3, y = x = 4, y = x = 5, y = 3 x = 6, y = 4 x = 7, y = 5 x = 0, y = 7 x =, y = 6 x = 4, y = 5 x = 6, y = 4 x = 8, y = 3 x = 0, y = teachitmaths.co.uk 03 6809
Simultaneous linear equations Solving two simultaneous equations involves finding the unique solution which satisfies both of them. There are two algebraic methods you could use: Elimination method go Substitution method go teachitmaths.co.uk 03 6809 3
Elimination method Comparing the coefficients between each equation tells us how difficult the simultaneous equations will be to solve. 6x + y = 0 3x + y = Beginner Matching coefficients go 3p + 4q = 4 p + 5q = 9 3x + 8y = 3 x + 7y = 3 Intermediate One coefficient is a multiple of the other Expert Coefficients are not multiples go go teachitmaths.co.uk 03 6809 4
Elimination method beginner 6x + y = 0 3x + y = 3x = 9 x = 3 3 3 + y = y =. Label the equations.. Compare the coefficients both y terms have a coefficient of. 3. Compare the signs both y terms are positive; subtract from to eliminate y. 4. Solve for x. 5. With an original equation, substitute x and solve for y. Why should we subtract, not add? teachitmaths.co.uk 03 6809 5
Elimination method beginner 6x + y = 0 3x + y = 3x = 9 x = 3 You can check your solution by substituting the values into both solutions: 6x + y = 0 6 3 + = 0 3 3 + y = y = 3x + y = 3 3 + = teachitmaths.co.uk 03 6809 6
Elimination method beginner 3p + q = 9 3p + 6q = 7 4q = 8 q = 3p + = 9 3p = 5 p = 5. Label the equations.. Compare the coefficients both p terms have a coefficient of 3. 3. Compare the signs both p terms are positive; subtract from to eliminate p. Why is this easier 4. Solve for q. than subtracting from? 5. With an original equation, substitute q and solve for p. teachitmaths.co.uk 03 6809 7
Elimination method beginner x + y = x y = 6 3x = 7 x = 9 9 + y = y = 3. Label the equations.. Compare the coefficients both y terms have a coefficient of. 3. Compare the signs each y sign is different; add and to eliminate y. 4. Solve for x. 5. Substitute x into an original equation and solve for y. Why should we add? teachitmaths.co.uk 03 6809 8
Elimination method beginner Solve these pairs of simultaneous equations: x + y = 30 x y = 4 4x + y = 3 3x + y = 8 x =, y = 4 x = 5, y = 3 x + y = 3x + y = 4 8k s = k s = x = 3, y = 5 k = 3, s = Click to show answers teachitmaths.co.uk 03 6809 9
Elimination method intermediate 3p + 4q = 4 p + 5q = 9. Label the equations.. Compare the coefficients no coefficients match. 3p + 4q = 4 3p + 5q = 57 q = 33 q = 3 3. Multiply all terms in by 3, to match the p coefficients. Why is it easier to eliminate p instead of q? 4. Compare the signs signs match; subtract. p + 5 3 = 9 p = 4 5. Solve for q. 6. Substitute q and solve for p. teachitmaths.co.uk 03 6809 0
Elimination method intermediate 5x + 4y = 3 3x y = 7 5x + 4y = 3 x 4y = 8 7x = 5 x = 3 3 3 y = 7 y =. Label the equations.. Compare the coefficients no coefficients match. 3. Multiply all terms in by 4, to match the y coefficients. 4. Compare the signs signs are different; add. 5. Solve for x. 6. Substitute x and solve for y. teachitmaths.co.uk 03 6809
Elimination method intermediate Solve these pairs of simultaneous equations: 3x + 7y = 47 x y = 3 3x + y = 39 5x + 7y = 3 4x + 9y = 3 3x + y = 4 x = 4, y = 5 x =, y = 4s + 3k = 39 s + 8k = 6 x = 0, y = 9 k =, s = 9 Click to show answers teachitmaths.co.uk 03 6809
Elimination method expert 3x + 8y = 3 x + 7y = 3 6x + 6y = 64 6x + y = 69 5y = 5 y = x + 7 = 3 x = 8. Label the equations.. Compare the coefficients no coefficients match. 3. Multiply by and by 3, to match the x coefficients. 4. Compare the signs signs match; subtract. 5. Solve for y. We could have chosen to make the y coefficients match instead how would you do this? 6. Substitute y and solve for x. teachitmaths.co.uk 03 6809 3
Elimination method expert x + 9y = 4 x + 6y =. Label the equations.. Compare the coefficients no coefficients match. x + 8y = 84 6x + 8y = 36 6x = 48 x = 3 3 + 6y = y = 3. Multiply by and by 3, to match the y coefficients. 4. Compare the signs signs match; subtract. 5. Solve for x. Why did we multiply by and 3 instead of 9 and 6? 6. Substitute x and solve for y. teachitmaths.co.uk 03 6809 4
Elimination method expert Solve these pairs of simultaneous equations: 9x + 3y = 30 6x y = 6 4x + 9y = 35 3x + 7y = 7 7x + y = 3 5x + 3y = 9 x = 3, y = x =, y = 8 8s + 4k = -6 3s + 6k = 3 x =, y = 3 k =, s = -3 Click to show answers teachitmaths.co.uk 03 6809 5
Substitution method If one variable can be expressed easily in terms of the other, you may prefer to use the substitution method. 6x + y = 9 4x + 3y = 5 y = 9 6x suitable for the substitution method 4x + 3y = 35 7x + 5y = 59 y = ⅓(35 4x) will be tricky with the substitution method teachitmaths.co.uk 03 6809 6
Substitution method 3x + y = 8 x y = 5 y = x 5 3x + (x 5) = 8 3x + 4x 0 = 8 x = 4 y = 4 5 y = 3. Label the equations.. Rewrite to give y in terms of x. 3. Substitute y into. 4. Solve for x. Remember to use brackets and watch out for signs! 5. Substitute x into an original equation and solve for y. teachitmaths.co.uk 03 6809 7
Substitution method x + 3y = 9 3x y = x = 9 3y 3(9 3y) y = 57 9y y = y = 5 x = 9 3 5 x = 4. Label the equations.. Rewrite to give x in terms of y. 3. Substitute x into. 4. Solve for y. 5. Substitute y into an original equation and solve for x. teachitmaths.co.uk 03 6809 8
Substitution method Solve these pairs of simultaneous equations using the substitution method: 3x + 5y = 9 x y = 4 3x + y = 3 5x + 7y = 7 4x + 9y = 3 3x + y = 4 x = 3, y = x =, y = s + 4k = 6 5s + 3k = 47 x = 4, y = k = -, s = 0 Click to show answers teachitmaths.co.uk 03 6809 9